Example Are A (4 , − 7 , 9) , B (6 , 4 , 4) and C (7 , 10 , − 6) the vertices of a right angle triangle? Solution. 2 3 1 − → , − → , − → AB = 11 AC = 17 BC = 6 − 5 − 15 − 10 ◮ − AB · − → → AC = 6 + 187 + 75 � = 0 . ◮ − BA · − → BC = ( −− → AB ) · − → → BC = − 2 − 66 − 50 � = 0 . ◮ − CA · − → CB = ( −− → AC ) · ( −− → BC ) = − → AC · − → → BC = 3 + 102 + 150 � = 0 .
Example Are A (4 , − 7 , 9) , B (6 , 4 , 4) and C (7 , 10 , − 6) the vertices of a right angle triangle? Solution. 2 3 1 − → , − → , − → AB = 11 AC = 17 BC = 6 − 5 − 15 − 10 ◮ − AB · − → → AC = 6 + 187 + 75 � = 0 . ◮ − BA · − → BC = ( −− → AB ) · − → → BC = − 2 − 66 − 50 � = 0 . ◮ − CA · − → CB = ( −− → AC ) · ( −− → BC ) = − → AC · − → → BC = 3 + 102 + 150 � = 0 . None of the angles is π 2 , and therefore the triangle is not a right angle triangle.
Example A rhombus is a parallelogram with sides of equal length. Prove that the diagonals of a rhombus are perpendicular.
Example A rhombus is a parallelogram with sides of equal length. Prove that the diagonals of a rhombus are perpendicular. Solution. Define the parallelogram (rhombus) by vectors � u and � v. � u Then the diagonals are � u + � v and � u − � v. v � Show that � u + � v and � u − � v are perpendicular.
Example A rhombus is a parallelogram with sides of equal length. Prove that the diagonals of a rhombus are perpendicular. Solution. Define the parallelogram (rhombus) by vectors � u and � v. � u Then the diagonals are � u + � v and � u − � v. v � Show that � u + � v and � u − � v are perpendicular. ( � u + � v ) · ( � u − � v ) = � u · � u − � u · � v + � v · � u + � v · � v u || 2 − � v || 2 = || � u · � v + � u · � v − || � u || 2 − || � v || 2 = || � 0 , since || � u || = || � v || . = Therefore, the diagonals are perpendicular.
is the projection of orthogonal, are and . Since , then Furthermore, if . for some , is parallel to Since . proj , written onto Projections u and � d , express � u as a sum � u = � u 1 + � u 2 , where � u 1 is Given nonzero vectors � parallel to � u 2 is orthogonal to � d and � d . � u � u u 2 � � u 1 � d
orthogonal, Since are and . Since , then Furthermore, if . for some , is parallel to Projections u and � d , express � u as a sum � u = � u 1 + � u 2 , where � u 1 is Given nonzero vectors � parallel to � u 2 is orthogonal to � d and � d . � u � u � u 2 � u 1 � d u onto � � u 1 is the projection of � d , written � u 1 = proj � d � u .
orthogonal, Projections u and � d , express � u as a sum � u = � u 1 + � u 2 , where � u 1 is Given nonzero vectors � parallel to � u 2 is orthogonal to � d and � d . � u � u � u 2 � u 1 � d u onto � � u 1 is the projection of � d , written � u 1 = proj � d � u . u 1 is parallel to � u 1 = t � d , � d for some t ∈ R . Since � Furthermore, if � u = � u 1 + � u 2 , then � u 2 = � u − � u 1 . Since � u 1 and � u 2 are
orthogonal, Projections u and � d , express � u as a sum � u = � u 1 + � u 2 , where � u 1 is Given nonzero vectors � parallel to � u 2 is orthogonal to � d and � d . � u � u � u 2 � u 1 � d u onto � � u 1 is the projection of � d , written � u 1 = proj � d � u . u 1 is parallel to � u 1 = t � d , � d for some t ∈ R . Since � Furthermore, if � u = � u 1 + � u 2 , then � u 2 = � u − � u 1 . Since � u 1 and � u 2 are � u 2 · � u 1 = 0 u 2 · ( t � � d ) = 0 u 2 · � t ( � d ) = 0 u 2 · � � d = 0
and therefore , we get Since u 2 · � � d = 0 u 1 ) · � ( � u − � d = 0 u · � u 1 · � d − � d = 0 � u · � d − ( t � d ) · � d = 0 � u · � d − t ( � d · � d ) = 0 � u · � d − t || � d || 2 = 0 � u · � t || � d || 2 � d =
and therefore u 2 · � � d = 0 u 1 ) · � ( � u − � d = 0 u · � u 1 · � d − � d = 0 � u · � d − ( t � d ) · � d = 0 � u · � d − t ( � d · � d ) = 0 � u · � d − t || � d || 2 = 0 � u · � t || � d || 2 � d = Since � d � = � 0 , we get u · � t = � d d || 2 , || �
and therefore u 2 · � � d = 0 u 1 ) · � ( � u − � d = 0 u · � u 1 · � d − � d = 0 � u · � d − ( t � d ) · � d = 0 � u · � d − t ( � d · � d ) = 0 � u · � d − t || � d || 2 = 0 � u · � t || � d || 2 � d = Since � d � = � 0 , we get u · � t = � d d || 2 , || � u · � d u 1 = � � � d . || � d || 2
Theorem u and � d be vectors with � d � = � Let � 0 .
Theorem u and � d be vectors with � d � = � Let � 0 . 1. u · � u = � d � proj � d � d . || � d || 2
Theorem u and � d be vectors with � d � = � Let � 0 . 1. u · � u = � d � proj � d � d . || � d || 2 2. u · � d u − � � � d || � d || 2 is orthogonal to � d.
Example 2 3 and � Let � u = − 1 v = 1 . Find vectors � u 1 and � u 2 so that 0 − 1 � u = � u 1 + � u 2 , with � u 1 parallel to � v and � u 2 orthogonal to � v.
Example 2 3 and � Let � u = − 1 v = 1 . Find vectors � u 1 and � u 2 so that 0 − 1 � u = � u 1 + � u 2 , with � u 1 parallel to � v and � u 2 orthogonal to � v. Solution. 3 15/11 u = � u · � v v = 5 = . u 1 = proj � 1 5/11 � v � v || 2 � || � 11 − 1 − 5/11
Example 2 3 and � Let � u = − 1 v = 1 . Find vectors � u 1 and � u 2 so that 0 − 1 � u = � u 1 + � u 2 , with � u 1 parallel to � v and � u 2 orthogonal to � v. Solution. 3 15/11 u = � u · � v v = 5 = . u 1 = proj � 1 5/11 � v � v || 2 � || � 11 − 1 − 5/11 2 3 7 7/11 − 5 = 1 = . � u 2 = � u − � u 1 = − 1 1 − 16 − 16/11 11 11 0 − 1 5 5/11
Distance from a Point to a Line Example Let P (3 , 2 , − 1) be a point in R 3 and L a line with equation x 2 3 = + t . y 1 − 1 z 3 − 2 Find the shortest distance from P to L, and find the point Q on L that is closest to P.
Distance from a Point to a Line Example Let P (3 , 2 , − 1) be a point in R 3 and L a line with equation x 2 3 = + t . y 1 − 1 z 3 − 2 Find the shortest distance from P to L, and find the point Q on L that is closest to P. Solution. Let P 0 = P 0 (2 , 1 , 3) be a point on L, � 3 P and let � − 2 � T . d = − 1 � u Then − − → − P 0 P, − − → 0 Q = − → 0 P 0 + − − → − → P 0 Q = proj � P 0 Q, d L P 0 and the shortest distance from P to L is the length of − QP, where − → QP = − → P 0 P − − − → − → P 0 Q. Q 0
Example (continued) − − → � T , � � T . � � P 0 P = 1 1 − 4 d = 3 − 1 − 2 − − → 3 15 P 0 P · � − − → − − → d d = 10 = 1 � . P 0 Q = proj � P 0 P = − 1 − 5 d || � 14 7 d || 2 − 2 − 10
Example (continued) − − → � T , � � T . � � P 0 P = 1 1 − 4 d = 3 − 1 − 2 − − → 3 15 P 0 P · � − − → − − → d d = 10 = 1 � . P 0 Q = proj � P 0 P = − 1 − 5 d || � 14 7 d || 2 − 2 − 10 Therefore, 2 15 29 − → + 1 = 1 , 0 Q = 1 − 5 2 7 7 3 − 10 11 � 29 7 , 2 7 , 11 � so Q = Q . 7
Example (continued) Finally, the shortest distance from P (3 , 2 , − 1) to L is the length of − → QP, where 1 15 − 4 QP = − − → P 0 P − − − → − → − 1 = 2 . P 0 Q = 1 − 5 6 7 7 − 4 − 10 − 9
Example (continued) Finally, the shortest distance from P (3 , 2 , − 1) to L is the length of − → QP, where 1 15 − 4 QP = − − → P 0 P − − − → − → − 1 = 2 . P 0 Q = 1 − 5 6 7 7 − 4 − 10 − 9 Therefore the shortest distance from P to L is √ ||− → QP || = 2 ( − 4) 2 + 6 2 + ( − 9) 2 = 2 � 133 . 7 7
and is a vector equation of the plane. Consider a plane containing a point or, equivalently, Then be an arbitrary point on this plane. , and let and orthogonal to vector Equations of Planes Given a point P 0 and a nonzero vector � n , there is a unique plane containing P 0 and orthogonal to � n . Definition A nonzero vector � n is a normal vector to a plane if and only if � n · � v = 0 for every vector � v in the plane.
and is a vector equation of the plane. be an arbitrary point on this plane. or, equivalently, Then Equations of Planes Given a point P 0 and a nonzero vector � n , there is a unique plane containing P 0 and orthogonal to � n . Definition A nonzero vector � n is a normal vector to a plane if and only if � n · � v = 0 for every vector � v in the plane. Consider a plane containing a point P 0 and orthogonal to vector � n , and let P
and is a vector equation of the plane. be an arbitrary point on this plane. or, equivalently, Then Equations of Planes Given a point P 0 and a nonzero vector � n , there is a unique plane containing P 0 and orthogonal to � n . Definition A nonzero vector � n is a normal vector to a plane if and only if � n · � v = 0 for every vector � v in the plane. Consider a plane containing a point P 0 and orthogonal to vector � n , and let P n · − − → � P 0 P = 0 ,
Then be an arbitrary point on this plane. or, equivalently, Equations of Planes Given a point P 0 and a nonzero vector � n , there is a unique plane containing P 0 and orthogonal to � n . Definition A nonzero vector � n is a normal vector to a plane if and only if � n · � v = 0 for every vector � v in the plane. Consider a plane containing a point P 0 and orthogonal to vector � n , and let P n · − − → � P 0 P = 0 , n · ( − 0 P − − → − → � 0 P 0 ) = 0 , and is a vector equation of the plane.
A scalar equation of the plane has the form The vector equation Now suppose is simply a scalar. where so Then the previous equation becomes . , and , where can also be written as n · ( − 0 P − − → − → � 0 P 0 ) = 0 n · − → n · − − → � 0 P = � 0 P 0 .
A scalar equation of the plane has the form The vector equation can also be written as is simply a scalar. where so Then the previous equation becomes where n · ( − 0 P − − → − → � 0 P 0 ) = 0 n · − → n · − − → � 0 P = � 0 P 0 . � T . Now suppose P 0 = P 0 ( x 0 , y 0 , z 0 ) , P = P ( x , y , z ) , and � n = � a b c
A scalar equation of the plane has the form The vector equation can also be written as is simply a scalar. where so Then the previous equation becomes where n · ( − 0 P − − → − → � 0 P 0 ) = 0 n · − → n · − − → � 0 P = � 0 P 0 . � T . Now suppose P 0 = P 0 ( x 0 , y 0 , z 0 ) , P = P ( x , y , z ) , and � n = � a b c a x a x 0 · = · , b y b y 0 c z c z 0
A scalar equation of the plane has the form The vector equation can also be written as so Then the previous equation becomes where n · ( − 0 P − − → − → � 0 P 0 ) = 0 n · − → n · − − → � 0 P = � 0 P 0 . � T . Now suppose P 0 = P 0 ( x 0 , y 0 , z 0 ) , P = P ( x , y , z ) , and � n = � a b c a x a x 0 · = · , b y b y 0 c z c z 0 ax + by + cz = ax 0 + by 0 + cz 0 , where d = ax 0 + by 0 + cz 0 is simply a scalar.
The vector equation can also be written as Then the previous equation becomes so n · ( − 0 P − − → − → � 0 P 0 ) = 0 n · − → n · − − → � 0 P = � 0 P 0 . � T . Now suppose P 0 = P 0 ( x 0 , y 0 , z 0 ) , P = P ( x , y , z ) , and � n = � a b c a x a x 0 · = · , b y b y 0 c z c z 0 ax + by + cz = ax 0 + by 0 + cz 0 , where d = ax 0 + by 0 + cz 0 is simply a scalar. A scalar equation of the plane has the form ax + by + cz = d , where a , b , c , d ∈ R .
Problem Find an equation of the plane containing P 0 (1 , − 1 , 0) and orthogonal to � T . � � n = − 3 5 2
Problem Find an equation of the plane containing P 0 (1 , − 1 , 0) and orthogonal to � T . � � n = − 3 5 2 Solution A vector equation of this plane is − 3 x − 1 · = 0 . 5 y + 1 2 z
Problem Find an equation of the plane containing P 0 (1 , − 1 , 0) and orthogonal to � T . � � n = − 3 5 2 Solution A vector equation of this plane is − 3 x − 1 · = 0 . 5 y + 1 2 z A scalar equation of this plane is − 3 x + 5 y + 2 z = − 3(1) + 5( − 1) + 2(0) = − 8 , i.e., the plane has scalar equation − 3 x + 5 y + 2 z = − 8 .
Here are two solutions to the problem of finding the shortest distance from a point to a plane.
Here are two solutions to the problem of finding the shortest distance from a point to a plane. Example Find the shortest distance from the point P (2 , 3 , 0) to the plane with equation 5 x + y + z = − 1 , and find the point Q on the plane that is closest to P.
Here are two solutions to the problem of finding the shortest distance from a point to a plane. Example Find the shortest distance from the point P (2 , 3 , 0) to the plane with equation 5 x + y + z = − 1 , and find the point Q on the plane that is closest to P. Solution 1 P (2 , 3 , 0) Pick an arbitrary point P 0 on the plane. � n Then − → − − → QP = proj � P 0 P, n ||− → QP || is the shortest distance, P 0 Q and − 0 Q = − → 0 P − − → → QP.
Here are two solutions to the problem of finding the shortest distance from a point to a plane. Example Find the shortest distance from the point P (2 , 3 , 0) to the plane with equation 5 x + y + z = − 1 , and find the point Q on the plane that is closest to P. Solution 1 P (2 , 3 , 0) Pick an arbitrary point P 0 on the plane. � n Then − → − − → QP = proj � P 0 P, n ||− → QP || is the shortest distance, P 0 Q and − 0 Q = − → 0 P − − → → QP. � T . � � n = 5 1 1
Here are two solutions to the problem of finding the shortest distance from a point to a plane. Example Find the shortest distance from the point P (2 , 3 , 0) to the plane with equation 5 x + y + z = − 1 , and find the point Q on the plane that is closest to P. Solution 1 P (2 , 3 , 0) Pick an arbitrary point P 0 on the plane. � n Then − → − − → QP = proj � P 0 P, n ||− → QP || is the shortest distance, P 0 Q and − 0 Q = − → 0 P − − → → QP. � T . Choose P 0 = P 0 (0 , 0 , − 1) . � � n = 5 1 1
Here are two solutions to the problem of finding the shortest distance from a point to a plane. Example Find the shortest distance from the point P (2 , 3 , 0) to the plane with equation 5 x + y + z = − 1 , and find the point Q on the plane that is closest to P. Solution 1 P (2 , 3 , 0) Pick an arbitrary point P 0 on the plane. � n Then − → − − → QP = proj � P 0 P, n ||− → QP || is the shortest distance, P 0 Q and − 0 Q = − → 0 P − − → → QP. � T . Choose P 0 = P 0 (0 , 0 , − 1) . � � n = 5 1 1 Then − − → � T . � P 0 P = 2 3 1
Example (continued) P (2 , 3 , 0) � 2 − − → 1 � T . � n P 0 P = 3 � 5 1 � T . � n = 1 P 0 Q
Example (continued) P (2 , 3 , 0) � 2 − − → 1 � T . � n P 0 P = 3 � 5 1 � T . � n = 1 P 0 Q − − → � 5 − → − − → P 0 P · � n n = 14 1 � T . 1 QP = proj � P 0 P = n || 2 � n || � 27
Example (continued) P (2 , 3 , 0) � 2 − − → 1 � T . � n P 0 P = 3 � 5 1 � T . � n = 1 P 0 Q − − → � 5 − → − − → P 0 P · � n n = 14 1 � T . 1 QP = proj � P 0 P = n || 2 � n || � 27 √ √ √ Since ||− → QP || = 14 27 = 14 3 , the shortest distance from P to the plane is 14 3 . 27 9 9
Example (continued) P (2 , 3 , 0) � 2 − − → 1 � T . � n P 0 P = 3 � 5 1 � T . � n = 1 P 0 Q − − → � 5 − → − − → P 0 P · � n n = 14 1 � T . 1 QP = proj � P 0 P = n || 2 � n || � 27 √ √ √ Since ||− → QP || = 14 27 = 14 3 , the shortest distance from P to the plane is 14 3 . 27 9 9 To find Q, we have � 2 � 5 − 0 Q = − → 0 P − − → → 0 � T − 14 1 � T 3 1 QP = 27 � − 16 1 − 14 � T . = 67 27
Example (continued) P (2 , 3 , 0) � 2 − − → 1 � T . � n P 0 P = 3 � 5 1 � T . � n = 1 P 0 Q − − → � 5 − → − − → P 0 P · � n n = 14 1 � T . 1 QP = proj � P 0 P = n || 2 � n || � 27 √ √ √ Since ||− → QP || = 14 27 = 14 3 , the shortest distance from P to the plane is 14 3 . 27 9 9 To find Q, we have � 2 � 5 − 0 Q = − → 0 P − − → → 0 � T − 14 1 � T 3 1 QP = 27 � − 16 1 − 14 � T . = 67 27 − 16 27 , 67 27 , − 14 � � Therefore Q = Q . 27
where is a vector that is orthogonal to both A mnemonic device: . and The Cross Product Definition � T and � � T . Then � � Let � u = x 1 y 1 z 1 v = x 2 y 2 z 2 y 1 z 2 − z 1 y 2 . � u × � v = − ( x 1 z 2 − z 1 x 2 ) x 1 y 2 − y 1 x 2
where A mnemonic device: The Cross Product Definition � T and � � T . Then � � Let � u = x 1 y 1 z 1 v = x 2 y 2 z 2 y 1 z 2 − z 1 y 2 . � u × � v = − ( x 1 z 2 − z 1 x 2 ) x 1 y 2 − y 1 x 2 Note. � u × � v is a vector that is orthogonal to both � u and � v .
A mnemonic device: The Cross Product Definition � T and � � T . Then � � Let � u = x 1 y 1 z 1 v = x 2 y 2 z 2 y 1 z 2 − z 1 y 2 . � u × � v = − ( x 1 z 2 − z 1 x 2 ) x 1 y 2 − y 1 x 2 Note. � u × � v is a vector that is orthogonal to both � u and � v . � � � i x 1 x 2 1 0 0 � � , where � ,� ,� � � � . � u × � v = j y 1 y 2 i = 0 j = 1 k = 0 � � � � � 0 0 1 k z 1 z 2 � �
Theorem w ∈ R 3 . Let � v , �
Theorem w ∈ R 3 . Let � v , � 1. � v × � w is orthogonal to both � v and � w.
Theorem w ∈ R 3 . Let � v , � 1. � v × � w is orthogonal to both � v and � w. w = � 2. If � v and � w are both nonzero, then � u × � 0 if and only if � v and � w are parallel.
Problem � T and � Find all vectors orthogonal to both � u = − 1 − 3 2 � T . (We previously solved this using the dot product.) � v = 0 1 1 �
Problem � T and � Find all vectors orthogonal to both � u = − 1 − 3 2 � T . (We previously solved this using the dot product.) � v = 0 1 1 � Solution � � i − 1 0 � − 5 � � � � � � i + � j − � . � u × � v = j − 3 1 = − 5 k = 1 � � � � � − 1 k 2 1 � � Any scalar multiple of � u × � v is also orthogonal to both � u and � v, so − 5 , t 1 ∀ t ∈ R , − 1 gives all vectors orthogonal to both � u and � v. (Compare this with our earlier answer.)
Distance between skew lines Problem Given two lines x 3 1 x 1 1 = + s = + t , L 1 : y 1 1 and L 2 : y 2 0 z − 1 − 1 z 0 2 A. Find the shortest distance between L 1 and L 2 . B. Find the shortest distance between L 1 and L 2 , and find the points P on L 1 and Q on L 2 that are closest together.
Distance between skew lines Problem Given two lines x 3 1 x 1 1 = + s = + t , L 1 : y 1 1 and L 2 : y 2 0 z − 1 − 1 z 0 2 A. Find the shortest distance between L 1 and L 2 . B. Find the shortest distance between L 1 and L 2 , and find the points P on L 1 and Q on L 2 that are closest together. Solution Choose P 1 (3 , 1 , − 1) on L 1 and P 2 (1 , 2 , 0) on L 2 . P P 1 1 1 Let � and � denote direction d 1 = 1 d 2 = 0 − 1 2 P 2 Q vectors for L 1 and L 2 , respectively.
. is and Therefore, the shortest distance between proj and proj and P P 1 (3 , 1 , − 1) 1 1 � , � d 1 = 1 d 2 = 0 − 1 2 The shortest distance between L 1 and L 2 is the length of the projection of − − − → P 2 (1 , 2 , 0) n = � d 1 × � Q P 1 P 2 onto � d 2 .
. and and proj Therefore, the shortest distance between and is proj P P 1 (3 , 1 , − 1) 1 1 � , � d 1 = 1 d 2 = 0 − 1 2 The shortest distance between L 1 and L 2 is the length of the projection of − − − → P 2 (1 , 2 , 0) n = � d 1 × � Q P 1 P 2 onto � d 2 . − 2 1 1 2 − − − → × = P 1 P 2 = 1 n = 1 0 − 3 � 1 − 1 2 − 1
. and Therefore, the shortest distance between and is and P P 1 (3 , 1 , − 1) 1 1 � , � d 1 = 1 d 2 = 0 − 1 2 The shortest distance between L 1 and L 2 is the length of the projection of − − − → P 2 (1 , 2 , 0) n = � d 1 × � Q P 1 P 2 onto � d 2 . − 2 1 1 2 − − − → × = P 1 P 2 = 1 n = 1 0 − 3 � 1 − 1 2 − 1 − − − → P 1 P 2 || = |− − − → − − − → P 1 P 2 · � n − − − → P 1 P 2 · � n | P 1 P 2 = � n , || proj � . proj � n n n || 2 || � || � n ||
and and P P 1 (3 , 1 , − 1) 1 1 � , � d 1 = 1 d 2 = 0 − 1 2 The shortest distance between L 1 and L 2 is the length of the projection of − − − → P 2 (1 , 2 , 0) n = � d 1 × � Q P 1 P 2 onto � d 2 . − 2 1 1 2 − − − → × = P 1 P 2 = 1 n = 1 0 − 3 � 1 − 1 2 − 1 − − − → P 1 P 2 || = |− − − → − − − → P 1 P 2 · � n − − − → P 1 P 2 · � n | P 1 P 2 = � n , || proj � . proj � n n n || 2 || � || � n || √ Therefore, the shortest distance between L 1 and L 2 is |− 8 | 14 = 4 14 . √ 7
Solution B. Now . Therefore, and This system has unique solution i.e., and , so and is orthogonal to both and 1 1 � , � d 1 = 1 d 2 = 0 ; − 1 2 P P 1 (3 , 1 , − 1) 3 + s − → 0 P = 1 + s for some s ∈ R ; − 1 − s P 2 (1 , 2 , 0) 1 + t − → Q 0 Q = 2 for some t ∈ R . 2 t
Solution B. and . Therefore, and This system has unique solution i.e., and 1 1 � , � d 1 = 1 d 2 = 0 ; − 1 2 P P 1 (3 , 1 , − 1) 3 + s − → 0 P = 1 + s for some s ∈ R ; − 1 − s P 2 (1 , 2 , 0) 1 + t − → Q 0 Q = 2 for some t ∈ R . 2 t � − 2 − s + t Now − → 1 + s + 2 t � T is orthogonal to both L 1 and L 2 , so PQ = 1 − s − → − → PQ · � PQ · � d 1 = 0 d 2 = 0 ,
Solution B. and . Therefore, and This system has unique solution i.e., and 1 1 � , � d 1 = 1 d 2 = 0 ; − 1 2 P P 1 (3 , 1 , − 1) 3 + s − → 0 P = 1 + s for some s ∈ R ; − 1 − s P 2 (1 , 2 , 0) 1 + t − → Q 0 Q = 2 for some t ∈ R . 2 t � − 2 − s + t Now − → 1 + s + 2 t � T is orthogonal to both L 1 and L 2 , so PQ = 1 − s − → − → PQ · � PQ · � d 1 = 0 d 2 = 0 , − 2 − 3 s − t = 0 s + 5 t = 0 .
Solution B. and Therefore, i.e., and 1 1 � , � d 1 = 1 d 2 = 0 ; − 1 2 P P 1 (3 , 1 , − 1) 3 + s − → 0 P = 1 + s for some s ∈ R ; − 1 − s P 2 (1 , 2 , 0) 1 + t − → Q 0 Q = 2 for some t ∈ R . 2 t � − 2 − s + t Now − → 1 + s + 2 t � T is orthogonal to both L 1 and L 2 , so PQ = 1 − s − → − → PQ · � PQ · � d 1 = 0 d 2 = 0 , − 2 − 3 s − t = 0 s + 5 t = 0 . This system has unique solution s = − 5 7 and t = 1 7 .
Solution B. i.e., and and 1 1 � , � d 1 = 1 d 2 = 0 ; − 1 2 P P 1 (3 , 1 , − 1) 3 + s − → 0 P = 1 + s for some s ∈ R ; − 1 − s P 2 (1 , 2 , 0) 1 + t − → Q 0 Q = 2 for some t ∈ R . 2 t � − 2 − s + t Now − → 1 + s + 2 t � T is orthogonal to both L 1 and L 2 , so PQ = 1 − s − → − → PQ · � PQ · � d 1 = 0 d 2 = 0 , − 2 − 3 s − t = 0 s + 5 t = 0 . This system has unique solution s = − 5 7 and t = 1 7 . Therefore, � 16 � 8 7 , 2 7 , − 2 � 7 , 2 , 2 � P = P Q = Q . 7 7
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